Answer
$${f_{xx}}\left( {x,y} \right) = - 6{y^3} + 6xy{\text{ and }}{f_{xy}}\left( {x,y} \right) = - 18x{y^2} + 3{x^2}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = - 3{x^2}{y^3} + {x^3}y \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 3{x^2}{y^3} + {x^3}y} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = - 3{y^3}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] + y\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr
& {\text{solving derivatives}} \cr
& {f_x}\left( {x,y} \right) = - 3{y^3}\left( {2x} \right) + y\left( {3{x^2}} \right) \cr
& {f_x}\left( {x,y} \right) = - 6x{y^3} + 3{x^2}y \cr
& \cr
& {\text{find }}{f_{xx}}\left( {x,y} \right) \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 6x{y^3} + 3{x^2}y} \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = - 6{y^3}\frac{\partial }{{\partial x}}\left[ x \right] + 3y\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = - 6{y^3} + 6xy \cr
& \cr
& {\text{find }}{f_{xy}}\left( {x,y} \right) \cr
& {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 6x{y^3} + 3{x^2}y} \right] \cr
& {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr
& {f_{xy}}\left( {x,y} \right) = - 18x{y^2} + 3{x^2} \cr} $$
